2017 AMC 10B Problem 7

The walk is exactly half the trip, so call that half-distance $d$. Then $d$ is the answer we want.

1. The speeds are in kilometers per hour, so change the time to hours too: $44$ minutes $= \frac{44}{60} = \frac{11}{15}$ hour.
2. Biking $d$ km at $17$ km per hour takes $\frac{d}{17}$ hours.
3. Walking $d$ km at $5$ km per hour takes $\frac{d}{5}$ hours.
4. The two times add up to the whole trip:
$$\frac{d}{17} + \frac{d}{5} = \frac{11}{15}.$$
5. Put the left side over the common denominator $85$: $\frac{5d}{85} + \frac{17d}{85} = \frac{22d}{85}$.
6. Solve: $d = \frac{11}{15} \cdot \frac{85}{22} = \frac{85}{30} = \frac{17}{6}$.
7. As a decimal, $\frac{17}{6} = 2.833\ldots$, which rounds to $2.8$.

The answer is $\boxed{\textbf{(C)}\ 2.8}$.

Two parts of a trip with the same length but different speeds give one equation: the times add up to the total. Choosing the unknown to be exactly what is asked (the walked half, not the whole trip) saves a step at the end. Keep all units matching: kilometers per hour with hours.

Split the time instead of the distance. Walking is slower: covering the same distance takes $\frac{17}{5}$ times as long on foot as on the bike. So the $44$ minutes split in the ratio $5 : 17$ (bike : walk). Since $5 + 17 = 22$ and $44 \div 22 = 2$, she biked $10$ minutes and walked $34$ minutes. In $34$ minutes at $5$ km per hour she covers $\frac{34}{60} \cdot 5 = \frac{17}{6} \approx 2.8$ km.